Total probability theorem:
Partition of Sample-Space:
Take a sample space 'S'. Let A1, A2, L An be the
set of the mutually exclusive and exhaustive set of sample space S.
These A1, A2, L, An events are said to partition the
Sample space in to 'n'-parts.
We have ![737_Total probability theorem1.png](https://www.expertsmind.com/../CMSImages/737_Total probability theorem1.png)
And ![2340_Total probability theorem2.png](https://www.expertsmind.com/../CMSImages/2340_Total probability theorem2.png)
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![2255_Total probability theorem.png](https://www.expertsmind.com/../CMSImages/2255_Total probability theorem.png)
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The total Probability Theorem:
Let 'A' be any event of S. We can write
are mutually exclusive,
would also be mutually exclusive.
![1742_Total probability theorem4.png](https://www.expertsmind.com/../CMSImages/1742_Total probability theorem4.png)
This is called as the total probability of the event A.
Remark:
- P(A/Ai) gives up contribution of Ai in occurrence of A.
Example: One bag has 4 white balls and 3 black balls and a 2nd bag contains 3 white balls and 5 black balls. One ball is drawn from 1st bag and placed unseen in the 2nd bag. What is the probability that a ball now drawn from 2nd bag is black ?
Sol: Bag - I Bag - II
4 w 3 w
3 B 5 B
Let A1 be the event that white ball is transferred from bag - I to bag - II and A2 be the event that black ball is transferred from bag - I to bag -II.
P (A1) =4/7, P(A2) = 3/7
Let 'A' be probability the finally a black ball is drawn from the 2nd bag P(A/A1) = 5/9, P(A/A2) = 6/9.
Now from the total probability theorem we get,
P(A) = P(A1) . P(A/A1) + P(A2) . P(A/A2) = ![57_Total probability theorem6.png](https://www.expertsmind.com/CMSImages/57_Total%20probability%20theorem6.png)
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